A bag contains 6 blue balls and 5 green balls and 2 red balls. A ball is chosen at random. What is the probability of getting a red ball?

Options

A. b + r = 9
B. br = 18
C. `|b - r| = 4`
D. `b//r = 2`

Answer

A, B

Explanation

Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls is
`p_(1) = (.^(r )C_(2))/(.^(r+b)C_(2)) = (r(r-1))/((r+b)(r+b-1))`
The probaility of drawing two blue balls is
`p_(2) = (.^(b)C_(2))/(.^(r+b)C_(2)) = (b(b-1))/((r+b)(r+b-1))`
The probability of drawing one red and one blue ball is
`P_(3) = (.^(r)C_(1)xx.^(b)C_(1))/(.^(r+b)C_(2))=(2br)/((r+b)(r+b-1))`
By hypothesis, `p_(1)=5p_(2)` and `p_(3) = 6p_(2)`. Therefore,
r(r-1) = 5b(b - 1) and 2br = 6b(b-1)
implies r = 6, b = 3